HDU 4856 Tunnels

Tunnels

Time Limit: 1500ms

Memory Limit: 32768KB

This problem will be judged on  HDU. Original ID: 4856
64-bit integer IO format: %I64d      Java class name: Main

 

Bob is travelling in Xi’an. He finds many secret tunnels beneath the city. In his eyes, the city is a grid. He can’t enter a grid with a barrier. In one minute, he can move into an adjacent grid with no barrier. Bob is full of curiosity and he wants to visit all of the secret tunnels beneath the city. To travel in a tunnel, he has to walk to the entrance of the tunnel and go out from the exit after a fabulous visit. He can choose where he starts and he will travel each of the tunnels once and only once. Now he wants to know, how long it will take him to visit all the tunnels (excluding the time when he is in the tunnels).

 

Input

The input contains mutiple testcases. Please process till EOF.
For each testcase, the first line contains two integers N (1 ≤ N ≤ 15), the side length of the square map and M (1 ≤ M ≤ 15), the number of tunnels.
The map of the city is given in the next N lines. Each line contains exactly N characters. Barrier is represented by “#” and empty grid is represented by “.”.
Then M lines follow. Each line consists of four integers x1, y1, x2, y2, indicating there is a tunnel with entrence in (x1, y1) and exit in (x2, y2). It’s guaranteed that (x1, y1) and (x2, y2) in the map are both empty grid.

 

Output

For each case, output a integer indicating the minimal time Bob will use in total to walk between tunnels.
If it is impossible for Bob to visit all the tunnels, output -1.

 

Sample Input

5 4

....#

...#.

.....

.....

.....

2 3 1 4

1 2 3 5

2 3 3 1

5 4 2 1

Sample Output

7

Source

2014西安全国邀请赛

 

解题：状压dp。1<<(j-1)表示第i条隧道被选。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int maxn = 20;

18 struct Tunnels {

19     int u,v,x,y;

20 };

21 Tunnels e[maxn];

22 int n,m,g[maxn][maxn],dp[1<<15][maxn];

23 char mp[maxn][maxn];

24 int bfs(Tunnels &a,const Tunnels &b) {

25     queue< pii >q;

26     static int vis[maxn][maxn];

27     static const int dir[4][2] = {0,-1,-1,0,1,0,0,1};

28     memset(vis,-1,sizeof(vis));

29     q.push(make_pair(a.x,a.y));

30     vis[a.x][a.y] = 0;

31     while(!q.empty()) {

32         pii now = q.front();

33         q.pop();

34         if(now.first == b.u && now.second == b.v)

35             return vis[now.first][now.second];

36         for(int i = 0; i < 4; i++) {

37             int x = now.first+dir[i][0];

38             int y = now.second+dir[i][1];

39             if(vis[x][y] > -1 || mp[x][y] == '#') continue;

40             vis[x][y] = vis[now.first][now.second]+1;

41             q.push(make_pair(x,y));

42         }

43     }

44     return -1;

45 }

46 int main() {

47     while(~scanf("%d %d",&n,&m)) {

48         memset(mp,'#',sizeof(mp));

49         for(int i = 1; i <= n; i++)

50             scanf("%s",mp[i]+1);

51         for(int i = 1; i <= m; i++)

52             scanf("%d %d %d %d",&e[i].u,&e[i].v,&e[i].x,&e[i].y);

53         for(int i = 1; i <= m; i++) {

54             for(int j = 1; j <= m; j++)

55                 g[i][j] = bfs(e[i],e[j]);

56         }

57         memset(dp,INF,sizeof(dp));

58         for(int i = 1; i <= m; i++) dp[1<<(i-1)][i] = 0;

59         int ans = INF;

60         for(int i = 1,M = 1 << m; i < M; i++) {

61             for(int j = 1; j <= m; j++) {

62                 if(i&(1<<(j-1))) {

63                     for(int k = 1; k <= m; k++) {

64                         if(k == j || (i&(1<<(k-1)) == 0) || g[k][j] == -1) continue;

65                         dp[i][j] = min(dp[i][j],dp[i^(1<<(j-1))][k]+g[k][j]);

66                     }

67                 }

68                 if(i == M-1) ans = min(ans,dp[i][j]);

69             }

70         }

71         printf("%d\n",ans == INF?-1:ans);

72     }

73     return 0;

74 }

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